From: israel@math.ubc.ca (Robert Israel)
Newsgroups: sci.math
Subject: Re: Eigenvalues and determinants.
Date: 26 Nov 1998 20:56:00 GMT
In article <365D8ECE.11863825@doc.ic.ac.uk>, ht97 writes:
|> Today I was told that the determinant of a matrix is equal to the
|> product of its eigenvalues. Could someone please supply a proof of
|> this?
Let P(t) = det(A - tI) be the characteristic polynomial of A.
For an n x n matrix this is a polynomial in t of degree n, and leading
term (-t)^n.
Now (r is an eigenvalue of A) <=> (A - r I) is not invertible
<=> P(r) = det(A - r I) = 0 <=> r - t is a factor of P(t).
In particular, if A has n distinct eigenvalues, then P(t) has n distinct
factors r - t, and since it has degree n and leading term (-t)^n,
P(t) must be the product of the factors r - t over all the eigenvalues r.
In particular, det(A) = P(0) is the product of the eigenvalues r.
If the eigenvalues are not distinct, the product must be taken with multiplicities. The multiplicity m(r) of eigenvalue r is the highest
power of r - t that divides P(t). We must have P(t) = product of
(r-t)^m(r) over all eigenvalues r (by the Fundamental Theorem of
Algebra every polynomial must factor into linear factors, and the only
possible linear factors are of this form), and then again
det(A) = P(0) = product of r^(m(r)).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
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From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Newsgroups: sci.math
Subject: Re: Eigenvalues and determinants.
Date: 26 Nov 1998 16:15:03 -0500
In article <365D8ECE.11863825@doc.ic.ac.uk>, ht97 wrote:
>Today I was told that the determinant of a matrix is equal to the
>product of its eigenvalues. Could someone please supply a proof of
>this?
Hard to accommodate you without knowing more about your background
knowledge. My path to proving this would be: find Schur decomposition of
the matrix:
A = U' * R * U, U unitary and R upper triangular,
plus:
U' is the inverse of U,
plus:
determinant of product is product of determinants.
We know the eigenvalues of an upper triangular matrix, and hence the
proof.
Cheers. ZVK(Slavek).